Generalized Weighted EP Elements in Banach Algebras

1. Introduction

Let $\mathcal{A}$ be a Banach algebra with involution *. An element $a\in \mathcal{A}$ has a group inverse if there exists an $x\in R$ such that

$$x{a}^2=a,a{x}^2=x,ax=xa.$$

Such an x is unique if it exists, is denoted by $a^{\#}$, and is called the group inverse of a (see [14]). As it is well known, a square complex matrix A has a group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$. An element a in $\mathcal{A}$ is an EP element if there exists an $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=ax=xa$ (see [19]). Evidently, $a\in \mathcal{A}$ is an EP element if and only if a has group inverse $a^{\#}$ and ${\left(a{a}^{\#}\right)}^{*}=a{a}^{\#}$ ( [9]). A square complex matrix A is an EP matrix if and only if $rank\left(A\right)=rank\left(A^{*}\right)$ (see [16]).

An element a in a Banach *-algebra $\mathcal{A}$ is a *-DMP element if there exist $m\in N$ and $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=xa=ax,a^m=x{a}^{m+1}.$ It was proved that $a\in \mathcal{A}$ is a *-DMP element if and only if $a^m\in \mathcal{A}$ is an EP element for some $m\in N$ (see [4,5,6]).

So as to extend the study of EP operators to the infinite-dimensional setting, the authors introduced the generalized EP element in a Banach *-algebra. An element $a\in \mathcal{A}$ is a generalized EP element if there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in \mathcal{A}isanEPelement,y\in {\mathcal{A}}^{qnil}.$$

Here, ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{{\displaystyle \frac{1}{n}}}=0\}.$

An element $a\in \mathcal{A}$ is Hermitian if $a=a^{*}$. Let $e,f\in \mathcal{A}$ be invertible Hermitian elements. Following Zhu and Wang, an element $a\in \mathcal{A}$ has $(e,f)$-MP inverse if there exists an $x\in \mathcal{A}$ such that

$$axa=a,xax=x,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa.$$

The preceding x is unique if it exists, and we denote it by $a_{e,f}^{†}$. The set of all $(e,f)$-MP invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_{e,f}^{†}$ (see [20]). An element a in $\mathcal{A}$ is an $(e,f)$-EP element if it has $(e,f)$-inverse and $a{a}_{e,f}^{†}=a_{e,f}^{†}a.$ We use ${\mathcal{A}}_{e,f}^{\mathsf{e}}$ to stand for the set of all $(e,f)$-EP elements in $\mathcal{A}$. For a pair of Hermitian positive definite matrices $(M,N)$, many results characterize when a complex matrix is an $(M,N)$-EP matrix in [16]. Evidently, $a\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ if and only if $a\in {\mathcal{A}}_{(e,f)}^{†}\bigcap {\mathcal{A}}^{\#}$ and $a_{(e,f)}^{†}=a^{\#}$(see [3]).

The motivation of this paper is to characterize when an element in a Banach algebra is the sum of an $(e,f)$-EP element and a quasi-nilpotent. We adopt:

Definition 1. 

An element $a\in \mathcal{A}$ is a generalized $(e,f)$-EP element if there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{qnil}.$$

In Section 2, we investigate elementary properties of generalized weighted EP elements in a Banach algebra. We proved that $a\in {\mathcal{A}}_{e,f}^{ⓔ}$ if and only if $a\in R_{e,f}^{†}$ and $a{a}_{e,f}^{†}=a_{e,f}^{†}a$.

In Section 3, we characterize generalized weighted EP element by using its polar-like property. We further proved that $a\in {\mathcal{A}}_{e,f}^{ⓔ}$ if and only if there exists $x\in comm\left(a\right)$ such that ${\left(eax\right)}^{*}=eax,{\left(fax\right)}^{*}=fax,a-axa\in {\mathcal{A}}^{qnil}.$

Definition 2. 

An element $a\in \mathcal{A}$ is an $(e,f)$-DMP element if there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{nil}.$$

Finally, in Section 4, we are concerned on weighted *-DMP elements. We establish that $a\in {\mathcal{A}}_{e,f}^{Ⓔ}$ is equivalent to the existence of $x\in \mathcal{A}$ and $n\in N$ with ${\left(eax\right)}^{*}=eax,{\left(fax\right)}^{*}=fax,a^n=x{a}^{n+1}$, which in turn is equivalent to the existence of $k\in N$ such that $a^k\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$. These results reveal new properties of *-DMP elements in a Banach algebra.

Throughout the paper, all Banach *-algebras are complex with an identity. An element p in $\mathcal{A}$ is a projection provided that $p^2=p=p^{*}$. The commutant of $a\in \mathcal{A}$ is defined by $comm\left(a\right)=\{x\in \mathcal{A}|xa=ax\}$.

2. Generalized $(e,f)$-EP Elements

The purpose of this section is to introduce a new weak inverse which is a natural generalization of EP element in a *-Banach algebra. Our starting points is the following.

Theorem 1. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ has $(e,f)$-EP decomposition.

(2)

There exists $x\in comm\left(a\right)$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-axa\in {\mathcal{A}}^{qnil}.$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}ey=y{f}^{-1}{z}^{*}=0,z\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{qnil}.$$

Set $x=z_{e,f}^{\mathsf{e}}$. One easily checks that

$$\begin{array}{ccc}\hfill xy& =\hfill & z_{e,f}^{\mathsf{e}}{e}^{-1}\left(ez{z}_{e,f}^{\mathsf{e}}\right)y=z_{e,f}^{\mathsf{e}}{e}^{-1}{\left(ez{z}_{e,f}^{\mathsf{e}}\right)}^{*}y\hfill \\ & =\hfill & z_{e,f}^{\mathsf{e}}{e}^{-1}{\left(z_{e,f}^{\mathsf{e}}\right)}^{*}\left(z^{*}ey\right)=0,\hfill \\ \hfill xa& =\hfill & z_{e,f}^{\mathsf{e}}z+z_{e,f}^{\mathsf{e}}y=z_{e,f}^{\mathsf{e}}z,\hfill \\ \hfill xax& =\hfill & z_{e,f}^{\mathsf{e}}z{z}_{e,f}^{\mathsf{e}}=z_{e,f}^{\mathsf{e}}=x,\hfill \end{array}$$

Moreover, we check that

$$\begin{array}{ccc}\hfill y{z}_{e,f}^{\mathsf{e}}& =\hfill & y{z}_{e,f}^{\mathsf{e}}z{z}_{e,f}^{\mathsf{e}}=y{f}^{-1}\left[f{z}_{e,f}^{\mathsf{e}}z\right]z_{e,f}^{\mathsf{e}}\hfill \\ & =\hfill & \left(y{f}^{-1}{z}^{*}\right){\left[f{z}_{e,f}^{\mathsf{e}}\right)}^{*}{z}_{e,f}^{\mathsf{e}}=0,\hfill \end{array}$$

and then

$$\begin{array}{ccc}\hfill ax& =\hfill & (z+y)z_{e,f}^{\mathsf{e}}=z{z}_{e,f}^{\mathsf{e}}+y{z}_{e,f}^{\mathsf{e}}\hfill \\ & =\hfill & z{z}_{e,f}^{\mathsf{e}}=z_{e,f}^{\mathsf{e}}z=xa.\hfill \end{array}$$

Then

$$\begin{array}{c}{\left(eax\right)}^{*}={\left(ez{z}_{e,f}^{\mathsf{e}}\right)}^{*}=ez{z}_{e,f}^{\mathsf{e}}=eax,\\ {\left(fxa\right)}^{*}={\left(f{z}_{e,f}^{\mathsf{e}}z\right)}^{*}=f{z}_{e,f}^{\mathsf{e}}z=fxa.\end{array}$$

Since $y{z}_{e,f}^{\mathsf{e}}=0$, we see that

$$a-axa=a(1-xa)=(y+z)(1-z_{e,f}^{\mathsf{e}}z)=y(1-z_{e,f}^{\mathsf{e}}z)=y\in {\mathcal{A}}^{qnil},$$

as desired.

$\left(2\right)\Rightarrow \left(1\right)$ By hypotheses, there exists $z\in comm\left(a\right)$ such that

$$z=zaz,{\left(eaz\right)}^{*}=eaz,{\left(fza\right)}^{*}=fza,a-aza\in {\mathcal{A}}^{qnil}.$$

Set $x=aza$ and $y=a-aza.$ Then $a=x+y$. We claim that $x\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$. Evidently, we verify that

$$\begin{array}{ccc}\hfill xzx& =\hfill & \left(aza\right)z\left(aza\right)=a\left(zaz\right)aza=a\left(zaz\right)a=aza=x,\hfill \\ \hfill zxz& =\hfill & z\left(aza\right)z=za\left(zaz\right)=zaz=z,\hfill \\ \hfill exz& =\hfill & eazaz=eaz,\hfill \\ \hfill fzx& =\hfill & fzaza=fza,\hfill \\ \hfill {\left(exz\right)}^{*}& =\hfill & {\left(eaz\right)}^{*}=eaz=exz,\hfill \\ \hfill {\left(fzx\right)}^{*}& =\hfill & {\left(fza\right)}^{*}=fza=fzx.\hfill \end{array}$$

Moreover, we have

$$xz=azaz=az=za=zazazx=zx.$$

Accordingly, $x\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ and $z=x_{e,f}^{\mathsf{e}}$.

Moreover, we see that

$$\begin{array}{ccc}\hfill x^{*}ey& =\hfill & {\left(aza\right)}^{*}e(1-az)a=a^{*}{\left(az\right)}^{*}{e}^{*}(1-az)a\hfill \\ & =\hfill & a^{*}{\left(eaz\right)}^{*}(1-az)a=0,\hfill \\ & =\hfill & a^{*}\left(eaz\right)(1-az)a=0.\hfill \end{array}$$

Since ${\left(fza\right)}^{*}=fza$, we have $f^{-1}{\left(za\right)}^{*}{f}^{*}=za$, and so $f^{-1}{\left(za\right)}^{*}=za{f}^{-1}$. This implies that ${\left(za{f}^{-1}\right)}^{*}=za{f}^{-1}$. Therefore

$$\begin{array}{ccc}\hfill y{f}^{-1}{x}^{*}& =\hfill & (a-aza)f^{-1}{\left(aza\right)}^{*}=a(1-za)f^{-1}{\left(aza\right)}^{*}\hfill \\ & =\hfill & a{\left[(1-za)f^{-1}\right]}^{*}{\left(aza\right)}^{*}=a{\left[\left(aza\right)(1-za)f^{-1}\right]}^{*}=0.\hfill \end{array}$$

Since $a-z{a}^2\in {\mathcal{A}}^{qnil}$, by using Cline’s formula, $y=a-aza\in {\mathcal{A}}^{qnil}$. Therefore we have a $(e,f)$-EP decomposition $a=x+y$, as asserted. □

Let $e\in {\mathcal{A}}^{-1}$ be Hermitian. Recall that an element $a\in \mathcal{A}$ has a generalized e-core inverse if there exists an $x\in R$ such that

$$x=a{x}^2,{\left(eax\right)}^{*}=eax,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Such an x is unique if it exists and we denote it by $a^{e,ⓓ}$ (see [2]).

Corollary 1. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ has $(e,f)$-EP decomposition.

(2)

There exists a unique $x\in comm\left(a\right)$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-axa\in {\mathcal{A}}^{qnil}.$$

Proof. 

$\left(2\right)\Rightarrow \left(1\right)$ This is obvious by Theorem 2.1.

$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.1, there exists a $x_1\in comm\left(a\right)$ such that

$$x_1=x_{1}a{x}_1,{\left(ea{x}_1\right)}^{*}=ea{x}_1,{\left(f{x}_{1}a\right)}^{*}=f{x}_{1}a,a-a{x}_{1}a\in {\mathcal{A}}^{qnil}.$$

Suppose that there exists a $x_2\in comm\left(a\right)$ such that

$$x_2=x_{2}a{x}_2,{\left(ea{x}_2\right)}^{*}=ea{x}_2,{\left(f{x}_{2}a\right)}^{*}=f{x}_{2}a,a-a{x}_{2}a\in {\mathcal{A}}^{qnil}.$$

Since $x_1\in comm\left(a\right)$, we have $x_1=a{x}_1^2$ and $\underset{n\to \infty }{lim}{\Vert a^n-x{a}^{n+1}\Vert }^{{\displaystyle \frac{1}{n}}}=0$. By virtue of [2], we see that $x_1=a^{e,ⓓ}=x_2$, as desired. □

We denote x in Corollary 2.2 by $a_{e,f}^{ⓔ}$, and call it the generalized $(e,f)$-EP inverse of a. We call a is a generalized $(e,f)$-EP element if a has generalized $(e,f)$-EP inverse. We use ${\mathcal{A}}_{e,f}^{ⓔ}$ to denote the set of all generalized $(e,f)$-EP elements in $\mathcal{A}$.

Theorem 2. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}^{e,ⓓ}\bigcap {\mathcal{A}}^{f,ⓓ}$ and $a^{e,ⓓ}=a^{f,ⓓ}\in comm\left(a\right).$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ As in the proof of Corollary 2.2, $x=a^{e,ⓓ}$. Likewise, $x=a^{f,ⓓ}$. In view of [2], $a^{e,ⓓ}=a^{f,ⓓ}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ This is obvious by Theorem 2.1 and [2]. □

Corollary 2. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}_{(f,e)}^{ⓔ}$.

In this case, $a_{e,f}^{ⓔ}=a_{(f,e)}^{ⓔ}.$

Proof. 

Straightforward from Theorem 2.3. □

An element $a\in R$ has generalized $(e,f)$-weighted Moore-Penrose inverse if there exists $x\in \mathcal{A}$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-axa\in R^{qnil}.$$

The preceding x is denoted by $a_{e,f}^{†}$. The set of all generalized $(e,f)$-weighted Moore-Penrose invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_{e,f}^{†}$. We now proceed to examine the generalized weighted EP property by using the generalized Moore-Penrose invertibility.

Theorem 3. 

Let $a\in R$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

$a\in R_{e,f}^{†}$ and $a{a}_{e,f}^{†}=a_{e,f}^{†}a$.

Proof. 

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{qnil}.$$

Then $x\in {\mathcal{A}}_{e,f}^{†}$ and $a_{e,f}^{†}=x_{e,f}^{†}$. Therefore $a\in R_{e,f}^{†}$ and $a{a}_{e,f}^{†}=a_{e,f}^{†}a$.

$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{qnil}.$$

Since $a{a}_{e,f}^{†}=a_{e,f}^{†}a$, we see that

$$\begin{array}{ccc}\hfill x{x}_{e,f}^{†}& =\hfill & x{x}_{e,f}^{†}+\left[y{f}^{-1}\right]\left[f{x}_{e,f}^{†}x\right]x_{e,f}^{†}\hfill \\ & =\hfill & (x+y)x_{e,f}^{†}=a{a}_{e,f}^{†}\hfill \\ & =\hfill & a_{e,f}^{†}a=x_{e,f}^{†}(x+y)\hfill \\ & =\hfill & x_{e,f}^{†}xx+\left[x_{e,f}^{†}{e}^{-1}\right]\left[ex{x}_{e,f}^{†}\right]y\hfill \\ & =\hfill & x_{e,f}^{†}x.\hfill \end{array}$$

This implies that $x\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$. Therefore $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. □

Recall that $a\in \mathcal{A}$ has generalized Drazin inverse if there exists an $x\in \mathcal{A}$ such that $a{x}^2=x,ax=xa,a-a^{2}x\in {\mathcal{A}}^{qnil}.$ Such an x is unique, if exists, and denote it by $a^d$.

Corollary 3. 

Let $a\in R$. Then the following are equivalent:

(1)

a is generalized $(e,f)$-EP.

(2)

$a\in R_{e,f}^{†}\bigcap {\mathcal{A}}^d$ and $a_{e,f}^{†}=a^d.$

Proof. 

$\left(2\right)\Rightarrow \left(1\right)$ In view of Theorem 2.3, $a\in {\mathcal{A}}^{e,ⓓ}$. Hence $a\in {\mathcal{A}}^d$. Set $x{=}_{e,f}^{†}$. As in the proof of Theorem 2.1, $a=axa+(a-axa),ax=xa,axa\in {\mathcal{A}}^{\#},a-axa\in {\mathcal{A}}^{qnil}$. Hence, $a^d={\left(axa\right)}^{\#}=x=a_{e,f}^{†},$ as required.

$\left(1\right)\Rightarrow \left(2\right)$ Since $a{a}^d=a^{d}a$, we have $a{a}_{e,f}^{ⓔ}=a_{e,f}^{ⓔ}a$. In light of Theorem 2.1, $a\in R_{e,f}^{ⓔ}$ and $a_{e,f}^{†}=a_{e,f}^{ⓔ},$ as asserted. □

3. Characterizations and Representations of Generalized Weighted EP-Inverses

The aim of this section is to characterize the generalized weighted EP-inverse and its representations are thereby presented.

Theorem 4. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{e,f}^{ⓔ}$. In view of Theorem 2.3, $a\in {\mathcal{A}}^{e,ⓓ}\bigcap {\mathcal{A}}^{f,ⓓ}$ and $x=a^{e,ⓓ}=a^{f,ⓓ}\in comm\left(a\right).$ By virtue of [2], $a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}^{e,\#}\bigcap {\mathcal{A}}^{f,\#}$. Moreover, we have

$$x={\left(a^d\right)}^2{\left(a^d\right)}^{e,\#}={\left(a^d\right)}^2{\left(a^d\right)}^{f,\#}.$$

Explicitly, we have ${\left(a^d\right)}^{e,\#}=a^{2}x={\left(a^d\right)}^{f,\#}$. Thus, ${\left(a^d\right)}_{e,f}^{†}=a^{2}x$. It is easy to verify that

$$a{\left(a^d\right)}_{e,f}^{†}=a\left(a^{2}x\right)=\left(a^{2}x\right)a={\left(a^d\right)}_{e,f}^{†}a.$$

In light of Theorem 2.5, $a^d$ has $(e,f)$-EP inverse and ${\left(a^d\right)}^{\mathsf{e}}=a^{2}x$. Accordingly, $a^d\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ and ${\left(a^d\right)}_{e,f}^{\mathsf{e}}={\left(a^d\right)}^{f,\#}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x={\left(a^d\right)}^2{\left(a^d\right)}_{e,f}^{\mathsf{e}}$. Then In view of Theorem 1.2, we have $a^d{\left(a^d\right)}_{e,f}^{†}={\left(a^d\right)}_{e,f}^{†}{a}^d.$ Hence $a^d\in {\mathcal{A}}_{e,f}^{\#}$ and ${\left(a^d\right)}_{e,f}^{\#}={\left(a^d\right)}_{e,f}^{\mathsf{e}}$. Thus, $a^d\in {\mathcal{A}}^{e,\#}\bigcap {\mathcal{A}}^{f,\#}.$ By virtue of [2], $a\in {\mathcal{A}}^{e,ⓓ}\bigcap {\mathcal{A}}^{f,ⓓ}$ and

$$a^{e,ⓓ}=x=a^{f,ⓓ}.$$

Since $a^d{\left(a^d\right)}_{e,f}^{\mathsf{e}}={\left(a^d\right)}_{e,f}^{\mathsf{e}}{a}^d$, we see that $a^2{a}^d{\left(a^d\right)}_{e,f}^{\mathsf{e}}={\left(a^d\right)}_{e,f}^{\mathsf{e}}{a}^2{a}^d$. Hence

$$ax=a{\left(a^d\right)}^2{\left(a^d\right)}_{e,f}^{\mathsf{e}}={\left(a^d\right)}^2{\left(a^d\right)}_{e,f}^{\mathsf{e}}a=xa.$$

According to Theorem 2.3,

$$a_{e,f}^{ⓔ}={\left(a^d\right)}^2{\left(a^d\right)}_{e,f}^{\mathsf{e}},$$

as asserted. □

Corollary 4. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}^{\mathsf{e}}$.

Proof. 

This is obvious by Theorem 2.7. □

We come now to characterize the generalized weighted EP elements by using the polar-like property.

Theorem 5. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

There exists an idempotent $p\in comm\left(a\right)$ such that $a+p\in {\mathcal{A}}^{-1},{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp$ and $ap\in {\mathcal{A}}^{qnil}$.

In this case, $a_{e,f}^{ⓔ}=(1-p){(a+p)}^{-1}.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.1, there exists $x\in comm\left(a\right)$ such that $a{x}^2=x,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-x{a}^2\in {\mathcal{A}}^{qnil}.$ Set $p=1-ax$. Then $p=p^2\in comm\left(a\right)$ and $a+p=a+1-ax$. We directly verify that ${(a+1-ax)}^{-1}=x+1-ax$. Then $a+p\in {\mathcal{A}}^{-1},{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp$ and $ap=a-a^{2}x\in {\mathcal{A}}^{qnil}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists an idempotent $p\in comm\left(a\right)$ such that $a+p\in {\mathcal{A}}^{-1},{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp$ and $ap\in {\mathcal{A}}^{qnil}$. Set $x=(1-p){(a+p)}^{-1}$. Then $x\in comm\left(a\right),x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa.$ Moreover, we have $a-axa=a-a^{2}x=a-a^2(1-p){(a+p)}^{-1}=a-{(a+p)}^2(1-p){(a-p)}^{-1}=a-(a+p)(1-p)=ap\in {\mathcal{A}}^{qnil}.$ Therefore $a_{e,f}^{ⓔ}=(1-p){(a+p)}^{-1},$ as required. □

Corollary 5. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}^d$ and $e{a}^{\pi },f{a}^{\pi }$ are projections.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ In view of Corollary 2.6, $a_{e,f}^{†}=a^d.$ Hence, $e{a}^{\pi }=e-ea{a}_{e,f}^{†}$. This implies that ${\left(e{a}^{\pi }\right)}^{*}=e^{*}-{\left(ea{a}_{e,f}^{†}\right)}^{*}=e-ea{a}_{e,f}^{†}=e{a}^{\pi }$. Likewise, ${\left(f{a}^{\pi }\right)}^{*}=f{a}^{\pi }$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, $a\in {\mathcal{A}}^d$ and $e{a}^{\pi },f{a}^{\pi }$ are projections. Choose $p=a^{\pi }$. Then $p^2=p\in comm\left(a\right)$ and $ep,fp$ are projections. Then $a+p\in {\mathcal{A}}^{-1}$ and $ap\in {\mathcal{A}}^{qnil}$. Therefore we complete the proof by Theorem 3.3. □

Corollary 6. 

Let $a,b\in {\mathcal{A}}_{e,f}^{ⓔ}$. If $ab=ba=0$, then $a+b\in {\mathcal{A}}_{e,f}^{ⓔ}$. In this case,

$${(a+b)}_{e,f}^{ⓔ}=a_{e,f}^{ⓔ}+b_{e,f}^{ⓔ}.$$

Proof. 

In view of Corollary 3.4, $a,b\in {\mathcal{A}}^d$ and $e{a}^{\pi },f{a}^{\pi },e{b}^{\pi }$ and $f{b}^{\pi }$ are projections. Hence $a+b\in {\mathcal{A}}^d$ and ${(a+b)}^d=a^d+b^d$. Moreover, ${(a+b)}^{\pi }=1-(a+b)(a^d+b^d)=a^{\pi }+b^{\pi }-1$. Therefore $e{(a+b)}^{\pi },f{(a+b)}^{\pi }$ are projections. This completes the proof by Corollary 3.4. □

As the following shows, the regularity condition on x in Theorem 2.1 can be dropped.

Theorem 6. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

There exists $x\in comm\left(a\right)$ such that

$${\left(eax\right)}^{*}=eax,{\left(fax\right)}^{*}=fax,a-axa\in {\mathcal{A}}^{qnil}.$$

In this case,

$$\begin{array}{ccc}\hfill a_{e,f}^{ⓔ}& =\hfill & p{(a+1-p)}^{-1},s=xax,q=as-{\left(as\right)}^2,\hfill \\ \hfill z& =\hfill & -{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^{\infty }}\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ k\end{array}\right){\left(4q{(4q-1)}^{-1}\right)}^k,p=as-(2as-1)z.\hfill \end{array}$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ This is trivial by Lemma 2.1.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in comm\left(a\right)$ such that

$${\left(eax\right)}^{*}=eax,{\left(fax\right)}^{*}=fax,a-axa\in {\mathcal{A}}^{qnil}.$$

Set $s=xax$. Then $s\in comm\left(a\right)$. We check that

$$\begin{array}{ccc}a-a^{2}s\hfill & =\hfill & a-axaxa\hfill \\ & =\hfill & (1+ax)(a-a^{2}x)\hfill \\ & \in \hfill & {\mathcal{A}}^{qnil},\hfill \\ s-s^{2}a\hfill & =\hfill & xax-xaxaxax\hfill \\ & =\hfill & x(a-a^{2}x)x+xax(a-a^{2}x)x\hfill \\ & \in \hfill & {\mathcal{A}}^{qnil}.\hfill \end{array}$$

$$as-{\left(as\right)}^2=(a-a^{2}s)s\in {\mathcal{A}}^{qnil}.$$

Let

$$q=as-{\left(as\right)}^2,z=-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^{\infty }}\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ k\end{array}\right){\left(4q{(4q-1)}^{-1}\right)}^k,p=as-(2as-1)z.$$

Thus we have $p^2=p$ and $az-p\in {\mathcal{A}}^{qnil}$. We easily check that $(a+1-as)((s+1-as)=1+(a-a^{2}s)(1-s)+(s-s^{2}a).$ Hence,

$$\begin{array}{ccc}a+1-p\hfill & =\hfill & (a+1-as)+(as-p)\in {\mathcal{A}}^{-1},\hfill \\ a(1-p)\hfill & =\hfill & (a-a^{2}s)+a(as-p)\in {\mathcal{A}}^{qnil}.\hfill \end{array}$$

Since $a\in comm\left(as\right)$, we have $pa=ap$. As ${\left(eax\right)}^{*}=eax$ and ${\left(fax\right)}^{*}=fax$, we see that ${\left(eas\right)}^{*}=eas$ and ${\left(fas\right)}^{*}=fas$. This implies that ${\left(eq\right)}^{*}={(eas-easas)}^{*}={\left(eas\right)}^{*}-{\left(as\right)}^{*}{\left(eas\right)}^{*}=eas-{\left(as\right)}^{*}eas=eas-{\left(eas\right)}^{*}as=eas-e{\left(as\right)}^2=eq$. Hence, ${\left(e(4q-1)\right)}^{*}=e(4q-1)$. This implies that ${(4q-1)}^{*}e{(4q-1)}^{-1}=e$, and then

$$e{(4q-1)}^{-1}={\left({(4q-1)}^{*}\right)}^{-1}e={\left(e{(4q-1)}^{*}\right)}^{-1}.$$

Thus we verify that

$$\begin{array}{ccc}\hfill {\left(eq{(4q-1)}^{-1}\right)}^{*}& =\hfill & {\left({(4q-1)}^{-1}\right)}^{*}{\left(eq\right)}^{*}\hfill \\ & =\hfill & {\left({(4q-1)}^{-1}\right)}^{*}eq={\left(e{(4q-1)}^{-1}\right)}^{*}q\hfill \\ & =\hfill & e{(4q-1)}^{-1}q=eq{(4q-1)}^{-1}.\hfill \end{array}$$

Therefore we have

$$\begin{array}{ccc}\hfill {\left(e{\left(q{(4q-1)}^{-1}\right)}^k\right)}^{*}& =\hfill & {\left({\left(q{(4q-1)}^{-1}\right)}^{k-1}\right)}^{*}e\left(q{(4q-1)}^{-1}\right)\hfill \\ & =\hfill & {\left(e{\left(q{(4q-1)}^{-1}\right)}^{k-1}\right)}^{*}\left(q{(4q-1)}^{-1}\right)\hfill \\ & =\hfill & {\left(e{\left(q{(4q-1)}^{-1}\right)}^{k-2}\right)}^{*}{\left(q{(4q-1)}^{-1}\right)}^2\hfill \\ & ⋮\hfill & \\ & =\hfill & e{\left(q{(4q-1)}^{-1}\right)}^k.\hfill \end{array}$$

Accordingly, we see that ${\left(ez\right)}^{*}=ez$, and so

$$\begin{array}{ccc}\hfill {\left(ep\right)}^{*}& =\hfill & {\left(eas-e(2as-1)z\right)}^{*}\hfill \\ & =\hfill & {\left(eas\right)}^{*}-z^{*}{\left(e(2as-1)\right)}^{*}\hfill \\ & =\hfill & eas-z^{*}\left(e(2as-1)\right)\hfill \\ & =\hfill & eas-{\left(ez\right)}^{*}(2as-1)=eas-ez(2as-1)=ep.\hfill \end{array}$$

Likewise, ${\left(fp\right)}^{*}=fp$. By virtue of Theorem 3.3, $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. □

Corollary 7. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{ⓔ}$.

(2)

There exists an element $x\in \mathcal{A}$ such that

$$x=a{x}^2,{\left(eax\right)}^{*}=exa,{\left(fax\right)}^{*}=fxa,a-axa\in {\mathcal{A}}^{qnil}.$$

In this case,

$$\begin{array}{ccc}\hfill a_{e,f}^{ⓔ}& =\hfill & p{(a+1-p)}^{-1},s=xax,q=as-{\left(as\right)}^2,\hfill \\ \hfill z& =\hfill & -{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^{\infty }}\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ k\end{array}\right){\left(4q{(4q-1)}^{-1}\right)}^k,p=as-(2as-1)z.\hfill \end{array}$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 2.1.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we verify that

$$\begin{array}{ccc}\hfill eax& =\hfill & {\left(exa\right)}^{*}={\left(xa\right)}^{*}e={\left(a{x}^{2}a\right)}^{*}e\hfill \\ & =\hfill & {\left[\left(ax\right)\left(xa\right)\right]}^{*}e={\left[\left(eax\right)\left(xa\right)\right]}^{*}\hfill \\ & =\hfill & {\left(xa\right)}^{*}{\left(eax\right)}^{*}={\left(xa\right)}^{*}\left(exa\right)\hfill \\ & =\hfill & {\left(exa\right)}^{*}xa=\left(eax\right)\left(xa\right)\hfill \\ & =\hfill & e\left(a{x}^2\right)a=exa.\hfill \end{array}$$

Since $e\in {\mathcal{A}}^{-1}$, we deduce that $ax=xa$. Therefore $x=a_{e,f}^{ⓔ}$, as asserted. □

Next, we characterize the generalized weighted EP element using the polar-like property of its generalized Drazin inverse.

Theorem 7. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{e,f}^{ⓔ}$ if and only if

(1)

$a\in {\mathcal{A}}^d$;

(2)

there exists an idempotent $p\in comm\left(a\right)$ such that

$$a^d-p\in {\mathcal{A}}^{-1},a^{d}p=0,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Proof. 

⟹ Let $x=a_{e,f}^{ⓔ}$. By virtue of Corollary 2.6, $a\in {\mathcal{A}}^d$ and $x=a^d$. Set $u=a^d-1+ax$ and $v=a^{2}x-1+ax$. Then $uv=vu$ and

$$uv=(a^d-1+a^2{a}^{d}x)(a^{2}x-1+a^2{a}^{d}x)=1.$$

Let $p=1-ax$. Then $a^d-p\in {\mathcal{A}}^{-1},p^2=p\in comm\left(a\right),{\left(ep\right)}^{*}=ep$ and ${\left(fp\right)}^{*}=fp$. Moreover, we have $a^{d}p=a^d(1-ax)=0$, as required.

⟸ By hypothesis, there exists an idempotent $p\in comm\left(a\right)$ such that

$$u:=a^d-p\in {\mathcal{A}}^{-1},a^{d}p=0,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Then $a^d(1-p)=u(1-p)$, and so $1-p=a^d(1-p)u^{-1}$. Set $x=(1-p){(a^d-p)}^{-1}$. Then

$$a^{d}x=a^d(1-p){(a^d-p)}^{-1}=(1-p)(a^d-p){(a^d-p)}^{-1}=1-p.$$

Then we verify that

$$a^d=a^{d}x{a}^d,x=x{a}^{d}x,{\left(e{a}^{d}x\right)}^{*}=e{a}^{d}x,{\left(f{a}^{d}x\right)}^{*}=f{a}^{d}x.$$

Then $a^d\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ by Theorem 1.2. According to Theorem 3.1, $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. □

Corollary 8. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{e,f}^{ⓔ}$ if and only if

(1)

$a\in {\mathcal{A}}^d$;

(2)

there exists an idempotent $p\in comm\left(a\right)$ and an invertible $u\in comm\left(a\right)$ such that

$$a^d=pu,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Proof. 

⟹ Obviously, $a\in {\mathcal{A}}^d$. By virtue of Theorem 3.8, there exists an idempotent $q\in comm\left(a\right)$ such that

$$u:=a^d-q\in {\mathcal{A}}^{-1},a^{d}q=0,{\left(eq\right)}^{*}=eq,{\left(fq\right)}^{*}=fq.$$

Then $a^d=a^d(1-q)=u(1-q)$. Set $p=1-q$. Then $a^d=pu,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp,$ as required.

$\left(2\right)⟸$ By hypothesis, there exists an idempotent $p\in comm\left(a\right)$ and an invertible $u\in comm\left(a\right)$ such that

$$a^d=pu,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Set $q=1-p$. Then $a^d-q=pu-(1-p)\in {\mathcal{A}}^{-1}$. Moreover, we have $a^{d}q=0,{\left(eq\right)}^{*}=ep,{\left(fq\right)}^{*}=fq.$ Therefore $a\in {\mathcal{A}}_{e,f}^{ⓔ}$ by Theorem 3.8. □

Corollary 9. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{ⓔ}$.

(2)

$a\in {\mathcal{A}}^d$ and there exists a projection $p\in comm\left(a\right)$ such that

$$a^d-p\in {\mathcal{A}}^{-1},a^{d}p=0.$$

(3)

$a\in {\mathcal{A}}^d$ and there exists an idempotent $p\in comm\left(a\right)$ and an invertible $u\in comm\left(a\right)$ such that

$$a^d=pu,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 3.8 and Corollary 3.9. □

4. Weighted *-DMP Elements

In this section, we are concerned with $(e,f)$-DMP elements in a Banach *-algebra. Many properties of *-DMP elements are thereby extended to the wider cases.

Lemma 1. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{Ⓔ}$.

(2)

There exist $x\in comm\left(a\right)$ and $n\in N$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a^n=x{a}^{n+1}.$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}ey=y{f}^{-1}{z}^{*}=0,z\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{nil}.$$

Set $x=z_{e,f}^{\mathsf{e}}$. As in the proof of Theorem 2.1, we have

$$xax=x,xa=ax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa.$$

Moreover, $a-axa=y\in {\mathcal{A}}^{nil}.$ Write ${(a-axa)}^n=0$ for some $n\in N$. Then $(1-ax)a^n=0$, and so $a^n=x{a}^{n+1}$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypotheses, there exist $z\in comm\left(a\right)$ and $n\in N$ such that

$$z=zaz,{\left(eaz\right)}^{*}=eaz,{\left(fza\right)}^{*}=fza,a^n=z{a}^{n+1}.$$

Then ${(a-axa)}^n=a^n-x{a}^{n+1}=0$, and so $a-axa\in {\mathcal{A}}^{nil}$. Set $x=aza$ and $y=a-aza.$ Then $a=x+y$. Analogous to the proof of Theorem 2.1, we prove that $x\in {\mathcal{A}}_{e,f}^{\mathsf{e}},x^{*}ey=0,y{f}^{-1}{x}^{*}=0$. Therefore $a\in {\mathcal{A}}_{e,f}^{Ⓔ}$. □

Lemma 2. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{e,f}^{Ⓔ}$ if and only if $a\in {\mathcal{A}}_{e,f}^{ⓔ}\bigcap {\mathcal{A}}^D$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Clearly, $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. In view of Lemma 4.1, there exist $x\in comm\left(a\right)$ and $n\in N$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a^n=x{a}^{n+1}.$$

Hence $x=a{x}^2,ax=xa,a^n=x{a}^{n+1}.$ Thus $a\in {\mathcal{A}}^D$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Let $x=a_{e,f}^{ⓔ}$. Then there exists some $x\in comm\left(a\right)$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-axa\in {\mathcal{A}}^{qnil}.$$

Let $z=axa$ and $y=a-axa$. As in the proof of Theorem 2.1, we have

$$a=z+y,z^{*}ey=y{f}^{-1}{z}^{*}=0,z\in {\mathcal{A}}_{e,f}^{\mathsf{e}},y\in {\mathcal{A}}^{qnil}.$$

In view of Corollary 2.6, $x=a^d=a^D$. Since $a\in {\mathcal{A}}^D$, we see that $a^m=a^D{a}^{m+1}=x{a}^{m+1}$ for some $m\in N$. Therefore we have

$$y^m={(a-axa)}^m={\left[(1-ax)a\right]}^m=(1-ax)a^m=a^m-x{a}^{m+1}=0.$$

Thus $y\in {\mathcal{A}}^{nil}$, and therefore $a\in {\mathcal{A}}_{e,f}^{Ⓔ}$. □

Theorem 8. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{Ⓔ}$.

(2)

There exist $x\in comm\left(a\right)$ and $n\in N$ such that

$$x=a{x}^2,{\left(eax\right)}^{*}=exa,{\left(fax\right)}^{*}=fxa,a^n=x{a}^{n+1}.$$

(3)

There exist $x\in comm\left(a\right)$ and $n\in N$ such that

$${\left(eax\right)}^{*}=eax,{\left(fax\right)}^{*}=fax,a^n=x{a}^{n+1}.$$

for some $n\in N$.

(4)

$a\in {\mathcal{A}}^D$ and $a^D\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{e,f}^{Ⓔ}$. Then the implication is true by Lemma 4.1.

$\left(2\right)\Rightarrow \left(3\right)$ This is trivial.

$\left(3\right)\Rightarrow \left(1\right)$ By virtue of Theorem 3.6, $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. Then $a\in {\mathcal{A}}^D$. Therefore $a\in {\mathcal{A}}_{e,f}^{Ⓔ}$ by Lemma 4.2.

$\left(1\right)\Rightarrow \left(4\right)$ By the argument above, $a\in {\mathcal{A}}^D$. In view of Theorem 3.1, $a^D\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$.

$\left(4\right)\Rightarrow \left(1\right)$ By virtue of Theorem 3.1, $a\in {\mathcal{A}}_{e,f}^{ⓔ}$. Therefore we obtain the result by Lemma 4.2. □

It is a well-established fact that an element a in $\mathcal{A}$ is *-DMP if and only if there exists a $k\in N$ such that $a^k$ is EP. We come now to derive a new characterizations of *-DMP elements.

Corollary 10. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ is *-DMP.

(2)

There exist $x\in \mathcal{A}$ and $n\in N$ such that

$$x=a{x}^2,{\left(ax\right)}^{*}=xa,a^n=x{a}^{n+1}.$$

(3)

There exist $x\in \mathcal{A}$ and $n\in N$ such that

$${\left(ax\right)}^{*}=ax,a^n=x{a}^{n+1}.$$

for some $n\in N$.

(4)

$a\in {\mathcal{A}}^D$ and $a^D\in \mathcal{A}$ is EP.

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 4.3. □

We now present the relationship between weighted *-DMP elements and weighted EP elements.

Theorem 9. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{Ⓔ}.$

(2)

There exists $m\in N$ such that $a^k\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ for any $k\ge m$.

(3)

$a^k\in {\mathcal{A}}_{e,f}^{\mathsf{e}}$ for some $k\in N$.

(4)

$a^k\in {\mathcal{A}}_{e,f}^{Ⓔ}$ for some $k\in N$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Lemma 4.1, there exist $x\in comm\left(a\right)$ and $m\in N$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a^m=x{a}^{m+1}.$$

Let $k\ge m$. Since $a{x}^2=x$, we have $a^k{x}^k=ax$. Then we verify that

$$\begin{array}{ccc}\hfill x^k{a}^k{x}^k& =\hfill & a{x}^{k+1}=(a{x}^m+1)x^{k-m}=x^m{x}^{k-m}=x^k,\hfill \\ \hfill {\left(e{a}^k{x}^k\right)}^{*}& =\hfill & e{a}^k{x}^k,{\left(f{x}^k{a}^k\right)}^{*}=f{x}^k{a}^k,\hfill \\ \hfill x^k{\left(a^k\right)}^2& =\hfill & a^{k+1}x=\left(x{a}^{m+1}\right)a^{k-m}=a^m{a}^{k-m}=a^k.\hfill \end{array}$$

This implies that ${\left(a^k\right)}_{e,f}^{\mathsf{e}}=x^k$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ This is trivial.

$\left(3\right)\Rightarrow \left(4\right)$ This is obvious.

$\left(4\right)\Rightarrow \left(1\right)$ In view of Theorem 4.3, $a^k\in {\mathcal{A}}^D$. Hence $a\in {\mathcal{A}}^D$. Let $p=1-a^k{\left(a^k\right)}_{e,f}^{Ⓔ}$. In light of Corollary 2.6, $p=1-a^k{\left(a^k\right)}^d=1-a{a}^d$. It follows by Theorem 3.8 that

$${\left(a^k\right)}^d-p\in {\mathcal{A}}^{-1},{\left(a^k\right)}^{d}p=0,{\left(ep\right)}^{*}=ep,{\left(fp\right)}^{*}=fp.$$

Since ${\left(a^k\right)}^d={\left(a^d\right)}^k$, we see that ${\left(a^d\right)}^k-p\in {\mathcal{A}}^{-1}$; hence, $a^d-p\in {\mathcal{A}}^{-1}$. Clearly, $a^{d}p=a^{k-1}\left[{\left(a^k\right)}^{d}p\right]=0$ and $p\in comm\left(a\right)$. According to Theorem 3.8, $a\in {\mathcal{A}}_{e,f}^{Ⓔ},$ as asserted. □

Corollary 11. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{Ⓔ}.$

(2)

There exists $m\in N$ such that $a^k\in {\mathcal{A}}^{\mathsf{e}}$ for any $k\ge m$.

(3)

$a^k\in {\mathcal{A}}^{\mathsf{e}}$ for some $k\in N$.

(4)

$a^k\in {\mathcal{A}}^{Ⓔ}$ for some $k\in N$.

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 4.5. □